Question

Question #fbf2a

Answer 1

This question may be missing some important info to proceed, so I made it up along the way. For my made-up volume of gastric acid you need about 4.5 tablets. The dosage would be 5, probably.

Explanation:

Let's first obtain the HCl molarity in the gastric "juice" solution.

`2.19g((HCl)/(36.463g)) = 0.0601mol = (0.0601mol)/L = 0.0601M`

This is the balanced neutralization reaction.

`3HCl(aq)+Al(OH)_3(s) to AlCl_3(aq)+3H_2O(l)`

Normally, there is some amount of solution to neutralize, here evidently none is given. I think this is almost needed in order to proceed (please correct me in the comments if not!), so I will make one up for example's sake. Let's say we have 1.50L of gastric "juice", a reasonable estimate.

`1.50L((0.0601mol)/L)((Al(OH)_3)/(3HCl))((78.01g)/(Al(OH)_3))((10^3mg)/g) = 2.34*10^3 mg`

Lastly, we'd then divide this mass of `Al(OH)_3` by the mass of one antacid tablet to arrive at our answer.

`(2.34*10^3 mg)/(520mg) = 4.5` tablets